An important reaction sequence in the industrial production of nitric acid is the following:
N₂(g) + 3H₂(g) $\rightarrow$ 2NH₃(g)
4NH₃(g) + 5O₂(g) $\rightarrow$ 4NO(g) + 6H₂O(l)
Starting from 20.0 mol of nitrogen gas in the first reaction, how many moles of oxygen gas are required in the second one?

Question

An important reaction sequence in the industrial production of nitric acid is the following:
N₂(g) + 3H₂(g)

\rightarrow

2NH₃(g)
4NH₃(g) + 5O₂(g)

\rightarrow

4NO(g) + 6H₂O(l)
Starting from 20.0 mol of nitrogen gas in the first reaction, how many moles of oxygen gas are required in the second one?

Quizplus · Accepted Answer

From the first reaction, 2 moles of NH3 react with 3 moles of HNO3 to produce 2 moles of NH₃. Therefore, 20.0 mol of nitrogen gas will produce 20.0/2 x 3 = 30.0 mol of HNO3. From the second reaction, 4 moles of NH₃ react with 5 moles of O2 to produce 4 moles of NO and 6 moles of water. Therefore, 30.0 mol of HNO3 will require 30.0/4 x 5 = 37.5 mol of O2. However, the stoichiometric coefficient of O2 is multiplied by a factor of 5/4 to balance the number of moles of NH₃ from the first reaction, resulting in a requirement of 37.5 x 5/4 = 46.9 mol of O2, which is rounded up to 50.0 mol. Therefore, the correct answer is D.

An Important Reaction Sequence in the Industrial Production of Nitric →\rightarrow→

An Important Reaction Sequence in the Industrial Production of Nitric $\rightarrow$